What is the general solution for cos^2(x)=(1-cos(2x))/2 ?

The general solution for cos^2(x)=(1-cos(2x))/2 is x=(3pi)/4+pin,x= pi/4+pin

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cos^2(x)=(1-cos(2x))/2

cos^{2} (x) = \frac{1 - cos ( 2 x )}{2}

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Solution steps

cos^{2} (x) = \frac{1 - cos ( 2 x )}{2}

Subtract

\frac{1 - cos ( 2 x )}{2}

from both sides

cos^{2} (x) - \frac{1 - cos ( 2 x )}{2} = 0

Simplify

cos^{2} (x) - \frac{1 - cos ( 2 x )}{2} : \frac{2 cos ^{2} ( x ) - 1 + cos ( 2 x )}{2}

\frac{2 cos ^{2} ( x ) - 1 + cos ( 2 x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos^{2} (x) - 1 + cos (2 x) = 0

Rewrite using trig identities

2 cos^{2} (x) - 2 sin^{2} (x) = 0

Factor

2 cos^{2} (x) - 2 sin^{2} (x) : 2 (cos (x) + sin (x)) (cos (x) - sin (x))

2 (cos (x) + sin (x)) (cos (x) - sin (x)) = 0

Solving each part separately

cos (x) + sin (x) = 0 or cos (x) - sin (x) = 0

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

cos^{2} (x) - 4 cos (x) - 1 = 0

- 9 sin (3 x) = 0

cos (4 x) - 7 cos (2 x) = 8

6 tan (B) + 14 = 0

- 4 tan^{2} (θ) - 4 tan (θ) - 9 = 8 tan (θ) - 4

What is the general solution for cos^2(x)=(1-cos(2x))/2 ?
The general solution for cos^2(x)=(1-cos(2x))/2 is x=(3pi)/4+pin,x= pi/4+pin